Ratio Test vs AST

cherry
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Homework Statement
Find the interval of convergence for the given power series.
Relevant Equations
N/A
Hi, I'm having difficulty understanding why the interval of convergence is (0, 18].
When I tested x=18, I got the following conclusion using the ratio test.
IMG_65D89D7F1999-1.jpeg


When I attempt using AST, the function still diverges as the lim (n -> inf) = 2^n / n ≠ 0.
What am I missing?

Thanks!
 

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cherry said:
Homework Statement: Find the interval of convergence for the given power series.
Relevant Equations: N/A

Hi, I'm having difficulty understanding why the interval of convergence is (0, 18].
When I tested x=18, I got the following conclusion using the ratio test.
View attachment 344579

When I attempt using AST, the function still diverges as the lim (n -> inf) = 2^n / n ≠ 0.
What am I missing?

Thanks!
The series is this:
$$\sum_{n = 1}^\infty \frac{(x - 9)^n}{n(-9)^n}$$
When you substitute x = 18, the numerator is not ##18^n##.
 
Mark44 said:
The series is this:
$$\sum_{n = 1}^\infty \frac{(x - 9)^n}{n(-9)^n}$$
When you substitute x = 18, the numerator is not ##18^n##.
Oh my gosh, thank you!
 
One of the most important things to learn from mathematics is to be careful about each step. It is definitely a learning process. Mathematics is one subject where you need to get a long string of steps correct to get the right answer, and it is fairly unique in that respect. Also, when you are thinking about the hard steps, it is often the easy ones where mistakes occur. So you should make it a habit to review your work with as much attention to the easy steps as you give to the hard steps.
 
It's a power series at the point ##a=9## whose radius of convergence ##R## is given by
<br /> \frac{1}{R} = \limsup _n \frac{1}{\sqrt[n]{n9^n}} = \frac{1}{9}.<br />
Hence, interval of convergence contains ##(0,18)##. For ##x=18## we get ## \sum \frac{(-1)^n}{n} ##, which converges. For ##x=0## we get divergence. So the interval of convergence is ##(0,18]##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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